How to calculate $y'(0)$ in system of implicit equation?
I'm given a system of implicit equations, namely
\begin{align*} 1. \quad &f_1 = x\cos z+y\cos x+z\cos y = 0\\ 2. \quad &f_2=\sin x + \sin y - \sin z = 0 \end{align*}
I've already proven via the implicit function theorem that these equations can be solved localy at $(0,0,0)$ for $y(x),z(x)$
Now I want to calculate $y'(0)$.
\begin{align*} 1. \quad &\frac{\partial f_1}{\partial x}=\cos z-xz'\sin z+y'\cos x-y\sin x+z'\cos y-zy'\sin y = 0\\ 2. \quad &\frac{\partial f_2}{\partial x}=\cos x + y'\cos y - z'\cos z = 0 \end{align*}
I suppose I can already plug in $x= 0$. Then...
\begin{align*} 1. \quad &\frac{\partial f_1}{\partial x}(0,y,z)=\cos z+y'+z'\cos y-zy'\sin y = 0 \Rightarrow y'=-\frac{\cos z+z'\cos y}{1-z\sin y}\\ 2. \quad &\frac{\partial f_2}{\partial x}(0,y,z)=1 + y'\cos y - z'\cos z = 0 \Rightarrow y'=\frac{z'\cos z-1}{\cos y} \end{align*} Is that OK so far? Do I need to do anything else?